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Bumek [7]
2 years ago
13

Nicholas and Elaine are planning to serve cheesecake for dessert at their wedding and have purchased twelve cheesecakes. If the

cheesecakes are divided evenly among the 125 wedding guests, how much cheesecake will each guest receive?
Mathematics
1 answer:
Svet_ta [14]2 years ago
6 0

12 cheesecakes

125 wedding guests

125 divided by 12=10.41666666666667=10.4

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Answer:

Step-by-step explanation:

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2 years ago
The equation y=2x represents a set of data. Which statement is not true?
Sonja [21]

Answer:

Option D is correct

The initial value is 2

Step-by-step explanation:

The equation of line passes through the origin is represented by:

y = mx where m is the slope or unit rate .

Direct Proportionality says that:

if y \propto x then the equation of the form is y = kx where k is the constant of proportionality.

Given the equation: y = 2x

Then by definition:

m = 2

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⇒ the unit rate is 2.

Also, the constant of proportionality is 2.

Therefore, the statement which is not true for the equation y=2x represents a set of data is: The initial value is 2

7 0
2 years ago
Read 2 more answers
A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

A' = 0

This gives:

0 = -512y^{-2} + 2

Collect Like Terms

512y^{-2} = 2

Multiply through by y^2

y^2 * 512y^{-2} = 2 * y^2

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Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

Length = 10

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Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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Answer:

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Which of the following equations have exactly one solution?
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Answer:

B.

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